By Raphael Salem
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Additional resources for Algebraic numbers and Fourier analysis (Heath mathematical monographs)
J. By X we denote a positive algebraic integer of the field of (), which we shall determine later. Ln-1 the conjugates of X. We have, x being a fixed point in X()mX (1) R = E(~) X«() - and m a rational integer > 0, I) (EOH + .. -) + R X«() - I) (EI()m-1 + E2()m-2 + ... + Em). Li(ai - l)aiP i=l =° (mod I). Llai n-l X«() - I)()p i=l . p. 2 }; 1 ~Iili I (mod 1). Let us now write (1), after breaking the sum in parenthesis into two parts, as (3) X()mx (Eft + ... ) + R l = X«() - I) = P+ Q+R. We have I Q I < X«() - (4) X ()-N-I I) 1 _ ()-I = ()N' We now choose X of the form X = Xl + X2() + ...
Suppose that the series (not vanishing identically) with Cn = 0 (~) represents a constant in each interval of CE. One can write Cn 'Yn . h = . WIt m 1"n ~ 0• It follows that the integrated series CoX - ~ 1"; einx Inl~l n represents a linear function in each interval of CE. Hence, the series is summable-R to zero in each interval of CE, and thus, by Theorem HA, converges to zero in each interval contiguous to E, the set E being, therefore, a set of multiplicity. Remark. If the series of the theorem represents a function of bounded variation, the series converging to zero in CE is a Fourier-Stieltjes series (in the usual terminology, the Fourier-Stieltjes series of a "measure" whose "support" is E).
Hence the power series PROOF. is bounded in the unit circle, and the nature of its coefficients shows that it is a polynomial, which proves the theorem in this case. Suppose now that a-I = O. Then fez) mayor may not have a pole inside the unit circle. The point f(O) = 11o is an interior point for the transformed domain. Let u = fez). To the circle C, I u - ao I < 0, in the u-plane there corresponds, for 0 small enough, a domain D in the z-plane, including the origin, and completely interior, say, to the circle I z I
Algebraic numbers and Fourier analysis (Heath mathematical monographs) by Raphael Salem